a)

Ta có:

\(A=x^2-2x-1=x^2-2x+1-2=\left(x-1\right)^2-2\)

\(\ge0-2=-2\)

Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)

b)\(B=4x^2+4x+8=4x^2+4x+1+7\)

\(=\left(2x+1\right)^2+7\ge0+7=7\)

Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)

c)

Ta có:

\(C=3x-x^2+2=2-\left(x^2-3x\right)\)

\(=2+\dfrac{9}{4}-\left(x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}\right)\)

\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)

Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

d) Ta có:

\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)

\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)

Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)

e) Ta có:

\(E=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

\(\ge0+0+2=2\)

Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

\(=-\left(x^2-4x-1\right)\)

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

<=> xaa ) C= x²-6x + 11= (x-3)² +2

ta co : (x-3)² + > hoặc = 2

=> C đạt giá trị nhỏ nhất khi C=2

<=> x=3

b) D =(x-1) (x+2)(x+3)(x+6)

= [ (x-1)(x+6)][(x+2)(x+3)]

=(x² +5x -6)(x²+5x +6)

=(x²+5x )² - 36

ta có (x² +5x)² -36 luôn > hoặc = -36

=> D đạt GTNN khi D = -36

<=>(x² + 5x)^{2 =0}

=> x = 0 hoac x =-5

c) E = x² - 4x + y² - 8y + 6

=(x² -4x +4 ) + (y² - 8y +16 ) -14

= (x -2)² +( y-4)² -14

ta co (x-2)² + (y-4)² -14 luôn > hoặc = -14

=> E dat GTNN khi E = -14

<=> (x-2)²​ =0 va (y-4)² =0

<=> x =2 va y=4

d) G =x² -4xy +5y² + 10x -22y + 28 ( de sai nha ban )

= [(x² - 4xy + 4y² ) + 10x -20y +25 ]+ ( y² -2y +1 ) +2

= [(x-2y)² + 10x - 20y + 25 ] + (y-1)² +2

= [( x-2y)² + 2. 5 (x-2y) + 25 ] + (y-1)² +2

= (x-2y +5)² + ( y-1)² +2

ta co (x-2y +5 )² + (y-1)² +2 luôn > hoặc = 0

=> G đạt GTNN khi (x-2y+5 )²=0 hoac (y-1)² =0

<=> y-1 = 0 => y = 1

,=> x =-3

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

viết lại đề đi

\(A=25x^2-20x+7\)

\(\Leftrightarrow A=\left(5x-2\right)^2+3\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow5x-2=0\Leftrightarrow x=\frac{2}{5}\)

Vậy \(minA=3\Leftrightarrow x=\frac{2}{5}\)

\(B=-x^2+2x-2\)

\(\Leftrightarrow B=-\left(x^2-2x+1\right)-3\)

\(\Leftrightarrow B=-\left(x-1\right)^2-3\le-3\)

Dấu " = " xảy ra \(\Leftrightarrow x=1\)

Vậy \(maxB=-3\Leftrightarrow x=1\)

\(C=9x^2-12x\)

\(\Leftrightarrow C=\left(9x^2-12x+4\right)-4\)

\(\Leftrightarrow C=\left(3x-2\right)^2-4\ge-4\)

Dấu " = " xảy ra \(\Leftrightarrow3x-2=0\Leftrightarrow x=\frac{2}{3}\)

Vậy \(minC=-4\Leftrightarrow x=\frac{2}{3}\)

\(D=3-10x^2-4xy-4y^2\)

\(\Leftrightarrow D=-\left(4y^2+4xy+x^2+9x^2\right)-3\)

\(\Leftrightarrow D=-\left[\left(2y-x\right)^2+3x^2\right]-3\le-3\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}2y-x=0\\3x^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=0\end{cases}}\)

Vậy \(maxD=-3\Leftrightarrow x=y=0\)

\(E=4x-x^2+1\)

\(\Leftrightarrow E=-\left(x^2-4x+4\right)+5\)

\(\Leftrightarrow E=-\left(x-2\right)^2+5\le5\)

Dấu " = " xảy ra \(\Leftrightarrow x=2\)

Vậy \(maxE=5\Leftrightarrow x=2\)

\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

\(A=\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-3\right)^2=0\)

\(\Leftrightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Vậy GTNN của \(A\) là \(2\) khi \(x=3\)

\(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

\(B=\left(x-10\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-10\right)^2=0\)

\(\Leftrightarrow\)\(x-10=0\)

\(\Leftrightarrow\)\(x=10\)

Vậy GTNN của \(B\) là \(1\) khi \(x=10\)

Chúc bạn học tốt ~ 

\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

\(A=\left(x-3\right)^2+2\)

Mà  \(\left(x-3\right)^2\ge0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(A_{Min}=2\Leftrightarrow x=3\)

b) \(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

\(B=\left(x-10\right)^2+1\)

Mà  \(\left(x-10\right)^2\ge0\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :  \(x-10=0\Leftrightarrow x=10\)

Vậy  \(B_{Min}=1\Leftrightarrow x=10\)

c)  \(C=x^2-4xy+5y^2+10x-22y+28\)

\(C=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(C=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\)\(\left(y^2-2y+1\right)+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

Mà  \(\left(x-2y+5\right)^2\ge0\)

      \(\left(y-1\right)^2\ge0\)

\(\Rightarrow C\ge2\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vây  \(C_{Min}=2\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

\(D=4x-x^2+3\)

\(-D=x^2-4x-3\)

\(-D=\left(x^2-4x+4\right)-7\)

\(-D=\left(x-2\right)^2-7\)

Mà  \(\left(x-2\right)^2\ge0\)

\(\Rightarrow-D\ge-7\)

\(\Leftrightarrow D\le7\)

Dấu "=" xảy ra khi :  \(x-2=0\Leftrightarrow x=2\)

Vậy  \(D_{Max}=7\Leftrightarrow x=2\)

\(E=-x^2+6x-11\)

\(-E=x^2-6x+11\)

\(-E=\left(x^2-6x+9\right)+2\)

\(-E=\left(x-3\right)^2+2\)

Mà  \(\left(x-3\right)^2\ge0\)

\(\Rightarrow-E\ge2\)

\(\Leftrightarrow E\le-2\)

Dấu "=" xảy ra khi  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(E_{Max}=-2\Leftrightarrow x=3\)